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中四數學基本三角學20分

發問:

解下列方程 1.sinθ-cosθ=0, 0度<θ<270度 2.3cosθ+4sinθ=0,0<θ<360 3.tanθ(3sinθ+2)=0,0≤ θ<360 4.4cosθ sinθ +cosθ =0,0≤ θ <360 5.sinθ -√3tanθ =0,0≤ θ <360 6.√2 sinθ =tanθ ,0≤ θ <360 求好心人幚手

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< async src="//pagead2.googlesyndication.com/pagead/js/adsbygoogle.js"> 1. sinθ-cosθ=0 for 0<θ<270 tanθ = 1 θ = 45 or 225 2. 3cosθ+4sinθ=0 for 0<θ<360 tanθ = -3/4 θ = 143 or 323 3. tanθ(3sinθ+2)=0 for 0≤θ<360 tanθ = 0 or sinθ = -2/3 θ = 0 or 180 for tanθ = 0 θ = 222 or 318 for sinθ = -2/3 Hence, θ = 0,180,222,318 4. 4cosθ sinθ +cosθ =0 for 0≤θ<360 cosθ = 0 or 4sinθ + 1 = 0 => sinθ = -1/4 θ = 90 or 270 for cosθ = 0 θ = 194 or 346 for sinθ = -1/4 Hence, θ = 90,194,270,346 5. sinθ -√3tanθ =0 for 0≤θ<360 sinθ = 0 or (1-√3/cosθ) = 0 for cosθ=/=0, i.e. θ=/=90,270 θ = 0 or 180 for sinθ = 0 1 - √3/cosθ = 0 => cosθ = √3 which has no real solution θ Hence, θ = 0 or 180 6. √2sinθ =tanθ for 0≤θ<360 sinθ (√2 - 1/cosθ) = 0 for cosθ=/= 0, i.e. θ=/=90,270 sinθ = 0 or cosθ = 1/√2 θ = 0 or 180 for sinθ = 0 θ = 45 or 315 for cosθ = 1/√2 Hence, θ = 0,45,180,315

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